Sum of power series using derivation or integration could anyone help with this question?
$$\sum_{n=1}^{\infty}\frac{(x-\frac{1}{2})^{n+1}}{n(n+1)}$$
I have to find sum of this power series using differentiation or integration.
Thanks a lot!
 A: Differentiate with respect to $x$ twice:
$$ S''(x) = \sum_{n=1}^{\infty} (x-1/2)^{n-1} = \sum_{k=0}^{\infty} (x-1/2)^{k}, $$
geometric series formula gives
$$ S''(x) = \frac{1}{1-(x-1/2)} = \frac{1}{3/2-x}. $$
Now you have to integrate this twice to get the answer, using (which you should check) $S(1/2)=S'(1/2)=0$.
A: Recall that:
$$\frac{\left(x-\frac{1}{2}\right)^{n+1}}{n(n+1)}=\int_{0}^{x-\frac{1}{2}}\frac{y^n}{n}dy.$$ 
Then:
$$\sum_{n=1}^{\infty}\frac{(x-\frac{1}{2})^{n+1}}{n(n+1)} = \sum_{n=1}^{\infty}\int_{0}^{x-\frac{1}{2}}\frac{y^n}{n}dy = \\ = \int_{0}^{x-\frac{1}{2}}\sum_{n=1}^{\infty}\frac{y^n}{n}dy = -\int_{0}^{x-\frac{1}{2}}\log(1-y)dy = \\
= -\left((y-1)(\log(1-y)-1)\right)_{0}^{x-\frac{1}{2}} = \\=-\left(\log\left(x-\frac{3}{2}\right)-1\right)\left(x - \frac{3}{2}\right)+1.$$
References:
http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions
A: Differentiate the sum $S(x)$ once reveals that 
$$S'(x)=\sum_{n=1}^{\infty}\, \frac{(x-1/2)^n}{n}$$
which we recognize as the series representation of $\log(1-(x-1/2))$.
Integrate once and apply the condition that $S(1/2)=0$.
